1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Factor 32x^5 + 243y^5

  1. Mar 12, 2012 #1
    Completely factor 32x^5 + 243y^5

    okay i used synthetic divison and i referenced a factorign calculator to get the answer


    i used f(x) = (2x)^5 + (3y)^5

    then f(-3y/2)= (2x)^5 + (3y)^5 = 0 (factor thereom)

    using synthetic divison and factor (-3y/2)

    i got.. 32x^4 -48x^3y + 72x^2y^2 -108xy^3 + 162y^4

    The answer is (2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4)

    i notice that the second peice (16x^4.......)

    the whole peice is actually half of what i long divided..

    so my problem is why am i not dividing to get that answer? and how can i put together what i knowÉ
  2. jcsd
  3. Mar 12, 2012 #2


    Staff: Mentor

    So what you have is
    (2x)5 + (3y)5 = (x - (-3y/2))(32x4 -48x3y + 72x2y2 -108xy3 + 162y4)
    = (x + 3y/2)(32x4 -48x3y + 72x2y2 -108xy3 + 162y4)

    If you pull a factor of 2 out of the long factor, and use it to multiply the first factor, you get what you show below.

  4. Mar 13, 2012 #3

    Just out of curiousity why am i allowed to divide a two out of the long factor, but i have to multiply it by the small factor? Like why is that allowed to happen, is there a rule of some sort... :O
  5. Mar 13, 2012 #4
    Maybe this will help clear the confusion up:

    (\frac{a}{2}+\frac{b}{2})(2a + 2b) &=& \frac{2}{2} (\frac{a}{2}+\frac{b}{2})(2a + 2b)\\
    &=& 2 \cdot (\frac{a}{2}+\frac{b}{2}) \cdot \frac{(2a + 2b)}{2}\\
    &=& (a+b)(a + b)

    Does that help clear up what was done above?
  6. Mar 13, 2012 #5
    So is this kind of a choice, of expanding the 2 within the first factor and using the 1/2 and targeting the second factor peicce with it.. argh im just havin a rough time with this idea..

    the onyl connection i can make (whichi 'm not sure is true) is it looks like you multiply the equation by (2/2) to satisfy both the top and bottom of the equation, and then the rules of math just do the rest... but that's just a guess..

    Can u explain to me why you multiply the equation by 2/2
  7. Mar 13, 2012 #6
    I was just illustrating why you can divide one term by 2 and multiply the other by 2. It's the same thing as multiplying by [itex]\frac{2}{2} = 1[/itex]. I just figured it was easier to see in a smaller example.
  8. Mar 13, 2012 #7
    i understand that connection and looking at the smaller example in terms of the math(that is very clear).

    I think i was just thinking about it to hard, or perhaps not in the right manner. thanks a gain for your help man.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook